Math 182                                                         Name:              Key

Exam 1                                                                        February 20, 2013

Show all work- Answers that are not fully justified will earn little or no credit

Be sure to include units in your answers where appropriate.

If a problem asks for an exact answer, then a decimal approximation is not sufficient (so don't round off p as 3.14, or 1/3 as .33, etc).  Other answers should be correct to at least 4 decimal places.

1.     Find the exact value of each of the following limits.  Note: You may use numerical and/or graphical means to check your answers, but these techniques will not be sufficient to justify that your answers are exact.

This is continuous at the point, so just evaluate:

This limit could also be found by algebraic means.

Letting x=0, we see A=3

Letting x=-1, we see -1C=2, or C=-2

Letting x=1, we see 4A+2B+C=4, so 12+2B-2=4, so B=-3

3.  The graph shown is y=f(t).  .

Find the following values.  You may approximate if

necessary, but must justify your answers.

a.

The area under the curve from 0 to 2 is about 10

(note that the shape is very nearly a trapezoid)

b.

c.  The x-values of g’s local maxima, if any.

g’ local extrema occur at x=6 and x=7.

At x=6 g’ changes from positive to negative, so g has a local maximum there.

4.  a.  Evaluate

b.  A particle moves along a straight line with a velocity of   v(t) = (t×sin t)  during the time 0 Ł t Ł 2p , where t is measured in seconds, and v(t) in feet/second.  Positive velocity corresponds to moving to the right, and negative velocity to the left.  At the end of the 2p seconds, where is the particle relative to where it started?

, so the particle is 2π feet to the left of its starting point.

5.  a.  Evaluate

b.  A tree sapling is initially 2 feet tall, and grows at a rate of feet/year.

Carefully INTERPRET in one or two complete sentences the significance of the answer to part (a) in this context.  Full credit can be earned for this interpretation even if you did not get the correct answer to part (a).

The definite integral of a rate of change of a function over some interval gives the total change of the original function over that interval.  In this case, the integral would represent the total amount of growth of this tree from the time it is planted until ∞ (forever).

So the tree will grow another 5 feet, to an ultimate height of 8 feet if it lives forever.

6.  Determine whether the following integrals converge or diverge.  For any that converge, find the value:

So it converges to 2.

:  handling the first of these integrals first (although we could start with either),  , so the integral diverges.

7.  The function shown is

a.  Draw on this graph the 2 rectangles used to approximate the integral of this function from -1 to 3 using the M2 method.  Then calculate (by hand) the value of the M2 approximation.

Value of M2 :       2*1+2*4=10                .

b.  Use your calculator to find the M10 and T10 approximations to this integral (round to the nearest thousandth).

Value of M10 :      10.786             .

Value of T10 :       10.889             .

c.  Recall the error bound for the trapezoid approximation:  .  Use this formula with an appropriately chosen value for K to determine the maximum possible error in the T10 approximation above.

(Hint: the derivative of a function of the form ).

So K can be chosen to be exactly 8ln(2)2 or some convenient number suitably close and larger than this –e.g. 4 would be a good choice.

Extra Credit:  Does the integral        converge or diverge? Carefully justify your answer.

Answer 1 (easiest):  , so the comparison test applies, and tells us that the desired integral must converge too, to a number between 0 and 1.

Answer 2 (longer):  Substitute so the integral becomes

.  The first of these terms is 0 (use L’hopitals rule, or simply notice that as b gets larger and larger, you are taking the ln of a number closer and closer to 1, so the ln would approach 0), so the final answer must be –ln(1/2) = ln(2).